Author

Topic: jf's Puzzles (Read 199745 times)


jf

Nice work. You are learning the tricks pretty fast.



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jf

:::::: New Puzzle Well, well, well. Pretty easy puzzles there, huh? Now I'll give you the real puzzle. I hope this one gives you all as much trouble as it gave me. Come one and all. Listen to her lovely voice. Direct from the City of Light, I present to you the beautiful, the one and only, La Chanteuse: nodolur



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Werbad

I'm going to assume that there is a way to deterministically solve it, without resorting to a statistical solution. Still have no clue how one would do that tho, this one's gonna take a while to figure out.



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Werbad

That was a tough one. Great puzzle! For La Chanteuse ( nodolur): punigevIt took me quite a while to figure out a way to do this, and if it wasn't for the F crate already on the matcher I might've still been sitting here trying to figure it out. The solution I came up with was basically: I take a random crate and generate a sequence of 16 crates using this and dropping it on the matcher, if I get a sequence of crates greater than the target value followed by a sequence of lower crates the first lower value will be the target value, assuming the crate picked was an F.
There are some false flags which are handled by taking the potential target, subtracting the random crate and checking if that value is greater than the target. If the random crate was an F the resulting crate will be Target+1, which will be greater in all cases but F, for which it's already solved. If the crate was anything other than F the result will be 0, which obviously won't ever be greater than the target.
If no false flag was triggered I deliver the crate to the matcher, twice to lock it in place since I don't halt the system.



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jf

That was a tough one. Great puzzle! For La Chanteuse ( nodolur): punigevGreat work, Werbad! It's one thing to figure out how it can be solved, and another thing to build the moving parts to do it. My solution works differently: I depend on a sequence of crates starting at 1 and counting up, instead of starting at F and counting down, and I only use a sequence of 15.The first crate is duplicated, which is harmless. The sequence I am hoping for then is 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F. Note that there is no zero crate. I drop these onto the matcher and divide the result into left crates, L, and right crates R.If the output is (a series of) L, followed by (a series of) R, then it must have started with a 1 crate, and the last of the L series is the target value.If the output is L, followed by R, followed by at least one more L, then it could not have started with a 1 crate.If the output is only L, and no R crates at all, then the results are inconclusive. It could be 0, 8, 0, 8, ... dropping on a 9, for example, or 4, 8, C, 0, 4, 8, ... dropping on a C. Or, the target crate is F.If the output is only R, and no L crates at all, then the target crate must be 0. Odd starting crates will drop sixteen nonzero values.If the output is R, followed by at least one L, then it did not start with a 1 crate.After coming up with this plan, the hard part was making a machine that would do this. It looks for either a sequence of L followed by a sequence of R, capturing the last L, or for 16 R values. Of course, in the case of a target F, it won't find either and remains solved from the start.For La Chanteuse ( nodolur): dakocar


« Last Edit: March 01, 2012, 06:00:00 PM by jf »

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Rene

For La Chanteuse ( nodolur): gosapurInteresting puzzle. As Werbad, it took me a while to find an algorithm that works as well. I was trying to figure out faster ways to solve it ( trying to reliably fabricate a 1 crate), but could not get them to complete for all values. Once I hit on this slower algorithm, the solution was actually not that hard. EDIT: I actually use the same algorithm as Werbad, but different solution


« Last Edit: March 01, 2012, 03:04:56 PM by Rene »

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jf

That's an elegant design, Rene. I haven't worked out exactly how the trap works  I'm guessing that the timing is such that a crate lands on the left platform under the double ++ only on the correct output stream, but haven't verified that detail yet. I also forgot to comment on your Ice Box solution dupubyv, which also has that clean aesthetic.



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Rene

That's an elegant design, Rene. I haven't worked out exactly how the trap works  I'm guessing that the timing is such that a crate lands on the left platform under the double ++ only on the correct output stream, but haven't verified that detail yet. I also forgot to comment on your Ice Box solution dupubyv, which also has that clean aesthetic. Thanks. The trap on the bottom right does all the matching work. The left dozer selects L crates and converts them into their original value, the right dozer selects R crates. The trap retains the first L crate. It holds it under the door key, if and only if it has been preceeded by at least one R crate, and is not followed by another R create. Otherwise, it pops up into the door key, and bocks the mechanism. That is, it recognizes the sequence: one or more R crates, followed by only L crates; exactly the sequence generated by an F initial crate.



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jf

*NEW PUZZLE* There was a slip of paper taped to it with "1 7 6 0" written on it. Other than that, I don't have anything more to tell you: Ring Ring! hanicim


« Last Edit: April 29, 2012, 10:18:33 AM by jf »

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jf

Well done! I wasn't expecting this puzzle to be solved so fast. Now let's see if you can find the solution in the dark: Ring Ring Ring kezirin



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Werbad

For Ring Ring Ring ( kezirin): cypodabHad to use a slightly slower approach there. I think there might be a way to handle the F and 7 crates that I ended up with in my other solution, but I didn't manage to find a solution without gates.



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jf

For Ring Ring Ring ( kezirin): cypodabHad to use a slightly slower approach there. I think there might be a way to handle the F and 7 crates that I ended up with in my other solution, but I didn't manage to find a solution without gates. Another nice solution  I was thinking of an intended solution using modular exponents and so I didn't even consider this possibility. My solution to Ring Ring and Ring Ring Ring: dazaharI started with x^phi(16) = x^12 = 1 in Z*16, and found that already x^4 = (x^2)^2 = 1, and all odd x mod 16 are in Z*16.


« Last Edit: April 29, 2012, 08:47:40 PM by jf »

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curly

For Ring Ring ( hanicim) : mubadyd (didn't use your mechanism)For Ring Ring Ring ( kezirin) : satubyd (tries all odd numbers until it finds the one that goes through the mechanism quickly)


« Last Edit: April 30, 2012, 01:29:55 AM by curly »

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